NJC_2025_H2_Physics_Prelim_P3_Ans

2025 SH2 H2 Physics Preliminary Examination Paper 3 9749/03 Section A 1(a) work done per unit mass in bringing a (small test) mass from infinity to that point B1 1(b)(i) correct read off of 𝜙 and x, e.g. B1 − GM R = −6.3 × 107 correct equation and substitution of G, M and R, e.g. B1 𝑀 = 6.3 × 107 × 6.4 × 106 6.67 × 10−11 = 6.0 × 1024 kg (shown) 1(b)(ii)1. attempt to apply Conservation of Energy in any understandable form, e.g. C1 Gain in KE = Loss in GP 1 2 𝑚𝑣2 = 𝑚(𝜙𝑖 − 𝜙𝑓) 𝑣 = √2(0 − (−2.1)) × 107 value of 𝜙𝑓 = −2.1 × 107 J kg−1 (read off when x = 3R) C1 = 6500 m s−1 A1 Comment: the most common mistake was substituting x = 2R. Some candidates thought the mass was in a circular orbit which is the wrong context. 1(b)(ii)2. 𝑎 = 𝐹 𝑚 = 𝐺𝑀𝑚 𝑥2 𝑚 = 𝐺𝑀 (3𝑅)2 = 6.67 × 10−11 × 6.0 × 1024 (3 × 6.4 × 106)2 allow also a = g (ecf awarded if sub x = 2R) C1 = 1.1 m s−2 A1 Comment: similar mistake to the previous part. It is inappropriate to find the gradient on the graph as this is an approximate method. The question asks you to calculate not estimate the acceleration. The more precise method should be chosen. Also, using equations of motion is wrong as the acceleration is not constant here. 1(b)(iii) lower (or more negative) gravitational potential energy due to presence of the Moon M1 hence higher speed A1

2(a)(i) change in momentum of molecule per collision = −mu − mu = −2mu OR = mu − (−mu) = 2mu A1 2(a)(ii) average force (= impulse per collision time between collisions) = 2mu 2x/u M1 = mu2 x (shown) 2(b)(i) W = pV = (7.0×105) [ (20 − 5) ×10−6 ] C1 = 10.5 J A1 Comment: some responses perhaps thought the work done is the area within the cycle. Negative sign should not be applied as this is work done by. 2(b)(ii) heating supplied to gas / J work done on gas / J increase in internal energy of gas / J 36.8 − 10.5 26.3 − 30.0 zero − 30.0 zero 3.7 3.7 third row (C → A) B1 second row (B → C) B1 first row (A → B) – value in red (allow ecf) B1 first row (A → B) – values in blue (allow ecf) B1 Comment: the sign for work done for A → B is sometimes given wrongly 2(b)(iii) useful work done = 10.5 − 3.7 = 6.8 J C1 efficiency = useful work done total energy input × 100% = 6.8 36.8 × 100% = 18% A1 Comment: few got this correct, in particular, the energy input must be the heat supplied to the system. No process can be 100% efficient or more.

3(a) When two or more waves meet at a point, B1 the resultant displacement at that point is equal to the vector sum of the displacements due to the individual waves at that point. B1 Examiner’s comment: The student did not score for this question. 3(b)(i) wavelength = 3.0 × 108 2.5 × 1010 B1 = 0.012 m 3(b)(ii)1. waves are in anti-phase at the sources and have no path difference to reach O M1 waves meet in anti-phase at